(1-y)(1+y2) =0
\(\Leftrightarrow1+y^2-y-y^3=0\)
\(\Leftrightarrow-y^3+y^2-y+1\)
\(\Leftrightarrow-y^2\left(y-1\right)-\left(y-1\right)=0\)
\(\Leftrightarrow\left(y-1\right)\left(-y^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y-1=0\\-y^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=1\\-y^2=1\end{matrix}\right.\)
vi -y2=1 (vo nghiem) nen \(S=\left\{1\right\}\)
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